Loss of mass in a fission reaction

Question

I was reading a textbook for fun this summer and I had difficulty to understand this uranium 235 fission reaction:

$$\ _0^1n + \ _{92}^{235}U \rightarrow \ _{38}^{94}Sr + \ _{54}^{139} Xe + 3 \ _0^1n$$

I count:

• 144 neutrons and 92 protons on the left hand side of the equation

• 144 neutrons and 92 protons on the right hand side of the equation

Since the mass of a proton is a constant ($m_P\approx1.67262 × 10^{−27} kg$), and idem for the mass of a neutron $(m_N \approx 1,675×10^{−27} kg)$, there should be exactly the same mass before and after the reaction (i.e. $144*m_N + 92*m_P$). So it seems $\Delta m = 0$.

Then where does the $\Delta m > 0$ come from when we compute $E = \Delta m\ c^2$? (Einstein's mass-energy relation)

Answer 1

Since the mass of a proton is a constant (mP≈1.67262×10−27kg), and idem for the mass of a neutron (mN≈1,675×10−27kg), there should be exactly the same mass before and after the reaction (i.e. 144∗mN+92∗mP).

Actually, this is incorrect. The mass of a free proton and the mass of a free neutron are fixed, but the mass of a nucleus is not equal to the sum of the masses of the protons and neutrons.

The mass of a free proton is 1.008 a.m.u. and the mass of a free neutron is 1.009 a.m.u., so the mass of the free nucleons is:

(92) (1.008 a.m.u.) + (143) (1.009 a.m.u.) = 236.959 a.m.u.

But the actual mass of a U-235 nucleus is 235.044 a.m.u. The difference between the sum of the free nucleon masses and the nucleus mass is called the mass deficit or (expressed in energy units) the binding energy. Because of the binding energy you cannot simply sum the number of nucleons to get the mass of a nucleus, you have to measure it.

The mass of Sr-94 is 93.915 a.m.u. and the mass of Xe-139 is 138.919 a.m.u. and two free neutrons are 1.009 a.m.u. each. So the total mass of the products is 234.851 a.m.u. That means that the mass of the U-235 nucleus is about 0.193 a.m.u. larger than the sum of the masses of the products.